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  <ul>
    <li><a href="#列表">列表</a></li>
    <li><a href="#实例">实例</a></li>
    <li><a href="#将列表当做堆栈使用">将列表当做堆栈使用</a></li>
    <li><a href="#实例-1">实例</a></li>
    <li><a href="#将列表当作队列使用">将列表当作队列使用</a></li>
    <li><a href="#实例-2">实例</a></li>
    <li><a href="#列表推导式">列表推导式</a></li>
    <li><a href="#实例-3">实例</a></li>
    <li><a href="#嵌套列表解析">嵌套列表解析</a></li>
    <li><a href="#del-语句">del 语句</a></li>
    <li><a href="#元组和序列">元组和序列</a></li>
    <li><a href="#集合">集合</a></li>
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    <article class="post">
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            <h1 class="post-title">Python3数据结构</h1>
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        <date class="post-meta meta-date">
            2021年9月15日
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            <h1 id="python3-数据结构">Python3 数据结构</h1>
<p>本章节我们主要结合前面所学的知识点来介绍Python数据结构。</p>
<hr>
<h2 id="列表">列表</h2>
<p>Python中列表是可变的，这是它区别于字符串和元组的最重要的特点，一句话概括即：列表可以修改，而字符串和元组不能。</p>
<p>以下是 Python 中列表的方法：</p>
<table>
<thead>
<tr>
<th style="text-align:left">方法</th>
<th style="text-align:left">描述</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:left">list.append(x)</td>
<td style="text-align:left">把一个元素添加到列表的结尾，相当于 a[len(a):] = [x]。</td>
</tr>
<tr>
<td style="text-align:left">list.extend(L)</td>
<td style="text-align:left">通过添加指定列表的所有元素来扩充列表，相当于 a[len(a):] = L。</td>
</tr>
<tr>
<td style="text-align:left">list.insert(i, x)</td>
<td style="text-align:left">在指定位置插入一个元素。第一个参数是准备插入到其前面的那个元素的索引，例如 a.insert(0, x) 会插入到整个列表之前，而 a.insert(len(a), x) 相当于 a.append(x) 。</td>
</tr>
<tr>
<td style="text-align:left">list.remove(x)</td>
<td style="text-align:left">删除列表中值为 x 的第一个元素。如果没有这样的元素，就会返回一个错误。</td>
</tr>
<tr>
<td style="text-align:left">list.pop([i])</td>
<td style="text-align:left">从列表的指定位置移除元素，并将其返回。如果没有指定索引，a.pop()返回最后一个元素。元素随即从列表中被移除。（方法中 i 两边的方括号表示这个参数是可选的，而不是要求你输入一对方括号，你会经常在 Python 库参考手册中遇到这样的标记。）</td>
</tr>
<tr>
<td style="text-align:left">list.clear()</td>
<td style="text-align:left">移除列表中的所有项，等于del a[:]。</td>
</tr>
<tr>
<td style="text-align:left">list.index(x)</td>
<td style="text-align:left">返回列表中第一个值为 x 的元素的索引。如果没有匹配的元素就会返回一个错误。</td>
</tr>
<tr>
<td style="text-align:left">list.count(x)</td>
<td style="text-align:left">返回 x 在列表中出现的次数。</td>
</tr>
<tr>
<td style="text-align:left">list.sort()</td>
<td style="text-align:left">对列表中的元素进行排序。</td>
</tr>
<tr>
<td style="text-align:left">list.reverse()</td>
<td style="text-align:left">倒排列表中的元素。</td>
</tr>
<tr>
<td style="text-align:left">list.copy()</td>
<td style="text-align:left">返回列表的浅复制，等于a[:]。</td>
</tr>
</tbody>
</table>
<p>下面示例演示了列表的大部分方法：</p>
<h2 id="实例">实例</h2>
<pre><code>&gt;&gt;&gt; a = [66.25, 333, 333, 1, 1234.5]
&gt;&gt;&gt; print(a.count(333), a.count(66.25), a.count('x'))
2 1 0
&gt;&gt;&gt; a.insert(2, -1)
&gt;&gt;&gt; a.append(333)
&gt;&gt;&gt; a
[66.25, 333, -1, 333, 1, 1234.5, 333]
&gt;&gt;&gt; a.index(333)
1
&gt;&gt;&gt; a.remove(333)
&gt;&gt;&gt; a
[66.25, -1, 333, 1, 1234.5, 333]
&gt;&gt;&gt; a.reverse()
&gt;&gt;&gt; a
[333, 1234.5, 1, 333, -1, 66.25]
&gt;&gt;&gt; a.sort()
&gt;&gt;&gt; a
[-1, 1, 66.25, 333, 333, 1234.5]
</code></pre><p>注意：类似 insert, remove 或 sort 等修改列表的方法没有返回值。</p>
<hr>
<h2 id="将列表当做堆栈使用">将列表当做堆栈使用</h2>
<p>列表方法使得列表可以很方便的作为一个堆栈来使用，堆栈作为特定的数据结构，最先进入的元素最后一个被释放（后进先出）。用 append() 方法可以把一个元素添加到堆栈顶。用不指定索引的 pop() 方法可以把一个元素从堆栈顶释放出来。例如：</p>
<h2 id="实例-1">实例</h2>
<pre><code>&gt;&gt;&gt; stack = [3, 4, 5]
&gt;&gt;&gt; stack.append(6)
&gt;&gt;&gt; stack.append(7)
&gt;&gt;&gt; stack
[3, 4, 5, 6, 7]
&gt;&gt;&gt; stack.pop()
7
&gt;&gt;&gt; stack
[3, 4, 5, 6]
&gt;&gt;&gt; stack.pop()
6
&gt;&gt;&gt; stack.pop()
5
&gt;&gt;&gt; stack
[3, 4]
</code></pre><hr>
<h2 id="将列表当作队列使用">将列表当作队列使用</h2>
<p>也可以把列表当做队列用，只是在队列里第一加入的元素，第一个取出来；但是拿列表用作这样的目的效率不高。在列表的最后添加或者弹出元素速度快，然而在列表里插入或者从头部弹出速度却不快（因为所有其他的元素都得一个一个地移动）。</p>
<h2 id="实例-2">实例</h2>
<pre><code>&gt;&gt;&gt; from collections import deque
&gt;&gt;&gt; queue = deque([&quot;Eric&quot;, &quot;John&quot;, &quot;Michael&quot;])
&gt;&gt;&gt; queue.append(&quot;Terry&quot;)           # Terry arrives
&gt;&gt;&gt; queue.append(&quot;Graham&quot;)          # Graham arrives
&gt;&gt;&gt; queue.popleft()                 # The first to arrive now leaves
'Eric'
&gt;&gt;&gt; queue.popleft()                 # The second to arrive now leaves
'John'
&gt;&gt;&gt; queue                           # Remaining queue in order of arrival
deque(['Michael', 'Terry', 'Graham'])
</code></pre><hr>
<h2 id="列表推导式">列表推导式</h2>
<p>列表推导式提供了从序列创建列表的简单途径。通常应用程序将一些操作应用于某个序列的每个元素，用其获得的结果作为生成新列表的元素，或者根据确定的判定条件创建子序列。</p>
<p>每个列表推导式都在 for 之后跟一个表达式，然后有零到多个 for 或 if 子句。返回结果是一个根据表达从其后的 for 和 if 上下文环境中生成出来的列表。如果希望表达式推导出一个元组，就必须使用括号。</p>
<p>这里我们将列表中每个数值乘三，获得一个新的列表：</p>
<pre><code>&gt;&gt;&gt; vec = [2, 4, 6]
&gt;&gt;&gt; [3*x for x in vec]
[6, 12, 18]
</code></pre><p>现在我们玩一点小花样：</p>
<pre><code>&gt;&gt;&gt; [[x, x**2] for x in vec]
[[2, 4], [4, 16], [6, 36]]
</code></pre><p>这里我们对序列里每一个元素逐个调用某方法：</p>
<h2 id="实例-3">实例</h2>
<pre><code>&gt;&gt;&gt; freshfruit = ['  banana', '  loganberry ', 'passion fruit  ']
&gt;&gt;&gt; [weapon.strip() for weapon in freshfruit]
['banana', 'loganberry', 'passion fruit']
</code></pre><p>我们可以用 if 子句作为过滤器：</p>
<pre><code>&gt;&gt;&gt; [3*x for x in vec if x &gt; 3]
[12, 18]
&gt;&gt;&gt; [3*x for x in vec if x &lt; 2]
[]
</code></pre><p>以下是一些关于循环和其它技巧的演示：</p>
<pre><code>&gt;&gt;&gt; vec1 = [2, 4, 6]
&gt;&gt;&gt; vec2 = [4, 3, -9]
&gt;&gt;&gt; [x*y for x in vec1 for y in vec2]
[8, 6, -18, 16, 12, -36, 24, 18, -54]
&gt;&gt;&gt; [x+y for x in vec1 for y in vec2]
[6, 5, -7, 8, 7, -5, 10, 9, -3]
&gt;&gt;&gt; [vec1[i]*vec2[i] for i in range(len(vec1))]
[8, 12, -54]
</code></pre><p>列表推导式可以使用复杂表达式或嵌套函数：</p>
<pre><code>&gt;&gt;&gt; [str(round(355/113, i)) for i in range(1, 6)]
['3.1', '3.14', '3.142', '3.1416', '3.14159']
</code></pre><hr>
<h2 id="嵌套列表解析">嵌套列表解析</h2>
<p>Python的列表还可以嵌套。</p>
<p>以下实例展示了3X4的矩阵列表：</p>
<pre><code>&gt;&gt;&gt; matrix = [
...     [1, 2, 3, 4],
...     [5, 6, 7, 8],
...     [9, 10, 11, 12],
... ]
</code></pre><p>以下实例将3X4的矩阵列表转换为4X3列表：</p>
<pre><code>&gt;&gt;&gt; [[row[i] for row in matrix] for i in range(4)]
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]
</code></pre><p>以下实例也可以使用以下方法来实现：</p>
<pre><code>&gt;&gt;&gt; transposed = []
&gt;&gt;&gt; for i in range(4):
...     transposed.append([row[i] for row in matrix])
...
&gt;&gt;&gt; transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]
</code></pre><p>另外一种实现方法：</p>
<pre><code>&gt;&gt;&gt; transposed = []
&gt;&gt;&gt; for i in range(4):
...     # the following 3 lines implement the nested listcomp
...     transposed_row = []
...     for row in matrix:
...         transposed_row.append(row[i])
...     transposed.append(transposed_row)
...
&gt;&gt;&gt; transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]
</code></pre><h2 id="del-语句">del 语句</h2>
<p>使用 del 语句可以从一个列表中依索引而不是值来删除一个元素。这与使用 pop() 返回一个值不同。可以用 del 语句从列表中删除一个切割，或清空整个列表（我们以前介绍的方法是给该切割赋一个空列表）。例如：</p>
<pre><code>&gt;&gt;&gt; a = [-1, 1, 66.25, 333, 333, 1234.5]
&gt;&gt;&gt; del a[0]
&gt;&gt;&gt; a
[1, 66.25, 333, 333, 1234.5]
&gt;&gt;&gt; del a[2:4]
&gt;&gt;&gt; a
[1, 66.25, 1234.5]
&gt;&gt;&gt; del a[:]
&gt;&gt;&gt; a
[]
</code></pre><p>也可以用 del 删除实体变量：</p>
<pre><code>&gt;&gt;&gt; del a
</code></pre><hr>
<h2 id="元组和序列">元组和序列</h2>
<p>元组由若干逗号分隔的值组成，例如：</p>
<pre><code>&gt;&gt;&gt; t = 12345, 54321, 'hello!'
&gt;&gt;&gt; t[0]
12345
&gt;&gt;&gt; t
(12345, 54321, 'hello!')
&gt;&gt;&gt; # Tuples may be nested:
... u = t, (1, 2, 3, 4, 5)
&gt;&gt;&gt; u
((12345, 54321, 'hello!'), (1, 2, 3, 4, 5))
</code></pre><p>如你所见，元组在输出时总是有括号的，以便于正确表达嵌套结构。在输入时可能有或没有括号， 不过括号通常是必须的（如果元组是更大的表达式的一部分）。</p>
<hr>
<h2 id="集合">集合</h2>
<p>集合是一个无序不重复元素的集。基本功能包括关系测试和消除重复元素。</p>
<p>可以用大括号({})创建集合。注意：如果要创建一个空集合，你必须用 set() 而不是 {} ；后者创建一个空的字典，下一节我们会介绍这个数据结构。</p>
<p>以下是一个简单的演示：</p>
<p>\</p>
<pre><code>&gt;&gt;&gt; basket = {'apple', 'orange', 'apple', 'pear', 'orange', 'banana'}
&gt;&gt;&gt; print(basket)                      # 删除重复的
{'orange', 'banana', 'pear', 'apple'}
&gt;&gt;&gt; 'orange' in basket                 # 检测成员
True
&gt;&gt;&gt; 'crabgrass' in basket
False

&gt;&gt;&gt; # 以下演示了两个集合的操作
...
&gt;&gt;&gt; a = set('abracadabra')
&gt;&gt;&gt; b = set('alacazam')
&gt;&gt;&gt; a                                  # a 中唯一的字母
{'a', 'r', 'b', 'c', 'd'}
&gt;&gt;&gt; a - b                              # 在 a 中的字母，但不在 b 中
{'r', 'd', 'b'}
&gt;&gt;&gt; a | b                              # 在 a 或 b 中的字母
{'a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'}
&gt;&gt;&gt; a &amp; b                              # 在 a 和 b 中都有的字母
{'a', 'c'}
&gt;&gt;&gt; a ^ b                              # 在 a 或 b 中的字母，但不同时在 a 和 b 中
{'r', 'd', 'b', 'm', 'z', 'l'}
</code></pre><p>集合也支持推导式：</p>
<pre><code>&gt;&gt;&gt; a = {x for x in 'abracadabra' if x not in 'abc'}
&gt;&gt;&gt; a
{'r', 'd'}

</code></pre><hr>
<h2 id="字典">字典</h2>
<p>另一个非常有用的 Python 内建数据类型是字典。</p>
<p>序列是以连续的整数为索引，与此不同的是，字典以关键字为索引，关键字可以是任意不可变类型，通常用字符串或数值。</p>
<p>理解字典的最佳方式是把它看做无序的键=&gt;值对集合。在同一个字典之内，关键字必须是互不相同。</p>
<p>一对大括号创建一个空的字典：{}。</p>
<p>这是一个字典运用的简单例子：</p>
<pre><code>&gt;&gt;&gt; tel = {'jack': 4098, 'sape': 4139}
&gt;&gt;&gt; tel['guido'] = 4127
&gt;&gt;&gt; tel
{'sape': 4139, 'guido': 4127, 'jack': 4098}
&gt;&gt;&gt; tel['jack']
4098
&gt;&gt;&gt; del tel['sape']
&gt;&gt;&gt; tel['irv'] = 4127
&gt;&gt;&gt; tel
{'guido': 4127, 'irv': 4127, 'jack': 4098}
&gt;&gt;&gt; list(tel.keys())
['irv', 'guido', 'jack']
&gt;&gt;&gt; sorted(tel.keys())
['guido', 'irv', 'jack']
&gt;&gt;&gt; 'guido' in tel
True
&gt;&gt;&gt; 'jack' not in tel
False
</code></pre><p>构造函数 dict() 直接从键值对元组列表中构建字典。如果有固定的模式，列表推导式指定特定的键值对：</p>
<pre><code>&gt;&gt;&gt; dict([('sape', 4139), ('guido', 4127), ('jack', 4098)])
{'sape': 4139, 'jack': 4098, 'guido': 4127}
</code></pre><p>此外，字典推导可以用来创建任意键和值的表达式词典：</p>
<pre><code>&gt;&gt;&gt; {x: x**2 for x in (2, 4, 6)}
{2: 4, 4: 16, 6: 36}
</code></pre><p>如果关键字只是简单的字符串，使用关键字参数指定键值对有时候更方便：</p>
<pre><code>&gt;&gt;&gt; dict(sape=4139, guido=4127, jack=4098)
{'sape': 4139, 'jack': 4098, 'guido': 4127}
</code></pre><hr>
<h2 id="遍历技巧">遍历技巧</h2>
<p>在字典中遍历时，关键字和对应的值可以使用 items() 方法同时解读出来：</p>
<pre><code>&gt;&gt;&gt; knights = {'gallahad': 'the pure', 'robin': 'the brave'}
&gt;&gt;&gt; for k, v in knights.items():
...     print(k, v)
...
gallahad the pure
robin the brave
</code></pre><p>在序列中遍历时，索引位置和对应值可以使用 enumerate() 函数同时得到：</p>
<pre><code>&gt;&gt;&gt; for i, v in enumerate(['tic', 'tac', 'toe']):
...     print(i, v)
...
0 tic
1 tac
2 toe
</code></pre><p>同时遍历两个或更多的序列，可以使用 zip() 组合：</p>
<pre><code>&gt;&gt;&gt; questions = ['name', 'quest', 'favorite color']
&gt;&gt;&gt; answers = ['lancelot', 'the holy grail', 'blue']
&gt;&gt;&gt; for q, a in zip(questions, answers):
...     print('What is your {0}?  It is {1}.'.format(q, a))
...
What is your name?  It is lancelot.
What is your quest?  It is the holy grail.
What is your favorite color?  It is blue.
</code></pre><p>要反向遍历一个序列，首先指定这个序列，然后调用 reversed() 函数：</p>
<pre><code>&gt;&gt;&gt; for i in reversed(range(1, 10, 2)):
...     print(i)
...
9
7
5
3
1
</code></pre><p>要按顺序遍历一个序列，使用 sorted() 函数返回一个已排序的序列，并不修改原值：</p>
<pre><code>&gt;&gt;&gt; basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']
&gt;&gt;&gt; for f in sorted(set(basket)):
...     print(f)
...
apple
banana
orange
pear
</code></pre>
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